IT小君
2021-09-15T00:20:37 我使用以下代码获取当前日期（格式为 12/31/1999 即 mm/dd/yyyy）：
Textview txtViewData;
txtViewDate.setText("Today is " +
android.text.format.DateFormat.getDateFormat(this).format(new Date()));
我还有另一个格式为：2010-08-25 (ie yyyy/mm/dd) 的日期，
所以我想找到日期之间的天数差异，我如何找到天数差异？
（换句话说，我想找到CURRENT DATE - yyyy/mm/dd 格式化日期之间的区别）
IT小君 这不是我的工作，在这里找到了答案。不希望将来断开链接:)。
关键是考虑到日光设置的这一行，参考完整代码。
TimeZone.setDefault(TimeZone.getTimeZone("Europe/London"));
或尝试通过TimeZone 作为参数来daysBetween()和呼叫setTimeZone()在sDate和eDate对象。
所以这里是这样的：
public static Calendar getDatePart(Date date){
Calendar cal = Calendar.getInstance(); // get calendar instance
cal.setTime(date);
cal.set(Calendar.HOUR_OF_DAY, 0); // set hour to midnight
cal.set(Calendar.MINUTE, 0); // set minute in hour
cal.set(Calendar.SECOND, 0); // set second in minute
cal.set(Calendar.MILLISECOND, 0); // set millisecond in second
return cal; // return the date part
}
getDatePart() 取自这里
/**
* This method also assumes endDate >= startDate
**/
public static long daysBetween(Date startDate, Date endDate) {
Calendar sDate = getDatePart(startDate);
Calendar eDate = getDatePart(endDate);
long daysBetween = 0;
while (sDate.before(eDate)) {
sDate.add(Calendar.DAY_OF_MONTH, 1);
daysBetween++;
}
return daysBetween;
}
细微差别：找出两个日期之间的差异并不像减去两个日期并将结果除以 (24 * 60 * 60 * 1000) 那样简单。事实上，它是错误的！
例如： 03/24/2007 和 03/25/2007 这两个日期之间的差应该是 1 天；但是，使用上述方法，在英国，您将获得0天！
自己看看（下面的代码）。以毫秒为单位将导致四舍五入错误，一旦您将夏令时之类的小东西纳入图片中，它们就会变得最明显。
完整代码：
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.TimeZone;
public class DateTest {
public class DateTest {
static SimpleDateFormat sdf = new SimpleDateFormat("dd-MMM-yyyy");
public static void main(String[] args) {
TimeZone.setDefault(TimeZone.getTimeZone("Europe/London"));
//diff between these 2 dates should be 1
Date d1 = new Date("01/01/2007 12:00:00");
Date d2 = new Date("01/02/2007 12:00:00");
//diff between these 2 dates should be 1
Date d3 = new Date("03/24/2007 12:00:00");
Date d4 = new Date("03/25/2007 12:00:00");
Calendar cal1 = Calendar.getInstance();cal1.setTime(d1);
Calendar cal2 = Calendar.getInstance();cal2.setTime(d2);
Calendar cal3 = Calendar.getInstance();cal3.setTime(d3);
Calendar cal4 = Calendar.getInstance();cal4.setTime(d4);
printOutput("Manual ", d1, d2, calculateDays(d1, d2));
printOutput("Calendar ", d1, d2, daysBetween(cal1, cal2));
System.out.println("---");
printOutput("Manual ", d3, d4, calculateDays(d3, d4));
printOutput("Calendar ", d3, d4, daysBetween(cal3, cal4));
}
private static void printOutput(String type, Date d1, Date d2, long result) {
System.out.println(type+ "- Days between: " + sdf.format(d1)
+ " and " + sdf.format(d2) + " is: " + result);
}
/** Manual Method - YIELDS INCORRECT RESULTS - DO NOT USE**/
/* This method is used to find the no of days between the given dates */
public static long calculateDays(Date dateEarly, Date dateLater) {
return (dateLater.getTime() - dateEarly.getTime()) / (24 * 60 * 60 * 1000);
}
/** Using Calendar - THE CORRECT WAY**/
public static long daysBetween(Date startDate, Date endDate) {
...
}
输出：
手册 - 2007 年 1 月 1 日和 2007 年 1 月 2 日之间的天数为：1
日历 - 2007 年 1 月 1 日和 2007 年 1 月 2 日之间的天数为：1
手册 - 2007 年 3 月 24 日和 2007 年 3 月 25 日之间的天数为：0
日历 - 2007 年 3 月 24 日和 2007 年 3 月 25 日之间的天数为：1
IT小君 大多数答案都很好，适合您的问题
所以我想找到日期之间的天数差异，我如何找到天数差异？
我建议采用这种非常简单直接的方法，可以保证在任何时区为您提供正确的差异：
int difference=
((int)((startDate.getTime()/(24*60*60*1000))
-(int)(endDate.getTime()/(24*60*60*1000))));
就是这样！
IT小君 Days.daysBetween(start.toDateMidnight() , end.toDateMidnight() ).getDays()
其中 'start' 和 'end' 是您的DateTime对象。要将日期字符串解析为 DateTime 对象，请使用parseDateTime 方法
IT小君 该片段考虑了夏令时并且是 O(1)。
private final static long MILLISECS_PER_DAY = 24 * 60 * 60 * 1000;
private static long getDateToLong(Date date) {
return Date.UTC(date.getYear(), date.getMonth(), date.getDate(), 0, 0, 0);
}
public static int getSignedDiffInDays(Date beginDate, Date endDate) {
long beginMS = getDateToLong(beginDate);
long endMS = getDateToLong(endDate);
long diff = (endMS - beginMS) / (MILLISECS_PER_DAY);
return (int)diff;
}
public static int getUnsignedDiffInDays(Date beginDate, Date endDate) {
return Math.abs(getSignedDiffInDays(beginDate, endDate));
}
这对我来说是简单和最好的计算，可能也适合你。
try {
/// String CurrDate= "10/6/2013";
/// String PrvvDate= "10/7/2013";
Date date1 = null;
Date date2 = null;
SimpleDateFormat df = new SimpleDateFormat("M/dd/yyyy");
date1 = df.parse(CurrDate);
date2 = df.parse(PrvvDate);
long diff = Math.abs(date1.getTime() - date2.getTime());
long diffDays = diff / (24 * 60 * 60 * 1000);
System.out.println(diffDays);
} catch (Exception e1) {
System.out.println("exception " + e1);
}
ChronoUnit.DAYS.between(
LocalDate.parse( "1999-12-28" ) ,
LocalDate.parse( "12/31/1999" , DateTimeFormatter.ofPattern( "MM/dd/yyyy" ) )
)
其他答案已过时。与最早版本的 Java 捆绑在一起的旧日期时间类已被证明设计不佳、混乱且麻烦。避开它们。
Joda-Time 项目非常成功地替代了那些旧类。这些类 为 Java 8 及更高版本中内置的java.time框架提供了灵感。
多的java.time功能后移植到Java 6和7在ThreeTen-反向移植和在进一步适于到Android ThreeTenABP。
LocalDate该LocalDate级表示没有时间一天和不同时区的日期，唯一的价值。
如果您的输入字符串采用标准ISO 8601格式，则LocalDate该类可以直接解析该字符串。
LocalDate start = LocalDate.parse( "1999-12-28" );
如果不是 ISO 8601 格式，请使用DateTimeFormatter.
String input = "12/31/1999";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern( "MM/dd/yyyy" );
LocalDate stop = LocalDate.parse( input , formatter );
ChronoUnit现在计算这对LocalDate对象之间经过的天数。该ChronoUnit枚举计算经过的时间。
long totalDays = ChronoUnit.DAYS.between( start , stop ) ;
如果您不熟悉 Java 枚举，请知道它们比大多数其他编程语言中的传统枚举更强大和有用。请参阅Enum课程文档、Oracle 教程和维基百科以了解更多信息。
该java.time框架是建立在Java 8和更高版本。这些类取代了麻烦的旧的遗留日期时间类，例如java.util.Date, Calendar, & SimpleDateFormat。
现在处于维护模式的Joda-Time项目建议迁移到java.time类。
要了解更多信息，请参阅Oracle 教程。并在 Stack Overflow 上搜索许多示例和解释。规范是JSR 310。
从哪里获得 java.time 类？
该ThreeTen-额外项目与其他类扩展java.time。该项目是未来可能添加到 java.time 的试验场。你可能在这里找到一些有用的类，比如Interval，YearWeek，YearQuarter，和更多。
IT小君 该Correct Way萨姆任务的回答只有当第一个日期比第二早期的作品。此外，如果两个日期在一天之内，它将返回 1。
这是最适合我的解决方案。就像大多数其他解决方案一样，由于错误的夏令时偏移，它仍然会在一年中的两天显示不正确的结果。
private final static long MILLISECS_PER_DAY = 24 * 60 * 60 * 1000;
long calculateDeltaInDays(Calendar a, Calendar b) {
// Optional: avoid cloning objects if it is the same day
if(a.get(Calendar.ERA) == b.get(Calendar.ERA)
&& a.get(Calendar.YEAR) == b.get(Calendar.YEAR)
&& a.get(Calendar.DAY_OF_YEAR) == b.get(Calendar.DAY_OF_YEAR)) {
return 0;
}
Calendar a2 = (Calendar) a.clone();
Calendar b2 = (Calendar) b.clone();
a2.set(Calendar.HOUR_OF_DAY, 0);
a2.set(Calendar.MINUTE, 0);
a2.set(Calendar.SECOND, 0);
a2.set(Calendar.MILLISECOND, 0);
b2.set(Calendar.HOUR_OF_DAY, 0);
b2.set(Calendar.MINUTE, 0);
b2.set(Calendar.SECOND, 0);
b2.set(Calendar.MILLISECOND, 0);
long diff = a2.getTimeInMillis() - b2.getTimeInMillis();
long days = diff / MILLISECS_PER_DAY;
return Math.abs(days);
}
最好和最简单的方法来做到这一点
public int getDays(String begin) throws ParseException {
long MILLIS_PER_DAY = 24 * 60 * 60 * 1000;
SimpleDateFormat dateFormat = new SimpleDateFormat("dd-MM-yyyy", Locale.ENGLISH);
long begin = dateFormat.parse(begin).getTime();
long end = new Date().getTime(); // 2nd date want to compare
long diff = (end - begin) / (MILLIS_PER_DAY);
return (int) diff;
}
使用以下功能：
/**
* Returns the number of days between two dates. The time part of the
* days is ignored in this calculation, so 2007-01-01 13:00 and 2007-01-02 05:00
* have one day inbetween.
*/
public static long daysBetween(Date firstDate, Date secondDate) {
// We only use the date part of the given dates
long firstSeconds = truncateToDate(firstDate).getTime()/1000;
long secondSeconds = truncateToDate(secondDate).getTime()/1000;
// Just taking the difference of the millis.
// These will not be exactly multiples of 24*60*60, since there
// might be daylight saving time somewhere inbetween. However, we can
// say that by adding a half day and rounding down afterwards, we always
// get the full days.
long difference = secondSeconds-firstSeconds;
// Adding half a day
if( difference >= 0 ) {
difference += SECONDS_PER_DAY/2; // plus half a day in seconds
} else {
difference -= SECONDS_PER_DAY/2; // minus half a day in seconds
}
// Rounding down to days
difference /= SECONDS_PER_DAY;
return difference;
}
/**
* Truncates a date to the date part alone.
*/
@SuppressWarnings("deprecation")
public static Date truncateToDate(Date d) {
if( d instanceof java.sql.Date ) {
return d; // java.sql.Date is already truncated to date. And raises an
// Exception if we try to set hours, minutes or seconds.
}
d = (Date)d.clone();
d.setHours(0);
d.setMinutes(0);
d.setSeconds(0);
d.setTime(((d.getTime()/1000)*1000));
return d;
}
有一个简单的解决方案，至少对我来说，是唯一可行的解决方案。
问题是我看到的所有答案都被扔了——使用 Joda、日历、日期或其他任何东西——只考虑了毫秒数。他们最终计算两个日期之间的 24 小时周期数，而不是实际天数。因此，从 1 月 1 日晚上 11 点到 1 月 2 日凌晨 1 点的时间将返回 0 天。
要计算startDate和之间的实际天数endDate，只需执行以下操作：
// Find the sequential day from a date, essentially resetting time to start of the day
long startDay = startDate.getTime() / 1000 / 60 / 60 / 24;
long endDay = endDate.getTime() / 1000 / 60 / 60 / 24;
// Find the difference, duh
long daysBetween = endDay - startDay;
这将在 1 月 2 日和 1 月 1 日之间返回“1”。如果您需要计算结束日期，只需将 1 添加到daysBetween（我需要在我的代码中这样做，因为我想计算范围内的总天数）。
这有点类似于Daniel 的建议，但我认为代码较小。
IT小君 所有这些解决方案都存在两个问题之一。由于舍入错误、闰日和秒数等原因，解决方案不完全准确，或者您最终会遍历两个未知日期之间的天数。
此解决方案解决了第一个问题，并将第二个问题提高了大约 365 倍，如果您知道最大范围是多少，效果会更好。
/**
* @param thisDate
* @param thatDate
* @param maxDays
* set to -1 to not set a max
* @returns number of days covered between thisDate and thatDate, inclusive, i.e., counting both
* thisDate and thatDate as an entire day. Will short out if the number of days exceeds
* or meets maxDays
*/
public static int daysCoveredByDates(Date thisDate, Date thatDate, int maxDays) {
//Check inputs
if (thisDate == null || thatDate == null) {
return -1;
}
//Set calendar objects
Calendar startCal = Calendar.getInstance();
Calendar endCal = Calendar.getInstance();
if (thisDate.before(thatDate)) {
startCal.setTime(thisDate);
endCal.setTime(thatDate);
}
else {
startCal.setTime(thatDate);
endCal.setTime(thisDate);
}
//Get years and dates of our times.
int startYear = startCal.get(Calendar.YEAR);
int endYear = endCal.get(Calendar.YEAR);
int startDay = startCal.get(Calendar.DAY_OF_YEAR);
int endDay = endCal.get(Calendar.DAY_OF_YEAR);
//Calculate the number of days between dates. Add up each year going by until we catch up to endDate.
while (startYear < endYear && maxDays >= 0 && endDay - startDay + 1 < maxDays) {
endDay += startCal.getActualMaximum(Calendar.DAY_OF_YEAR); //adds the number of days in the year startDate is currently in
++startYear;
startCal.set(Calendar.YEAR, startYear); //reup the year
}
int days = endDay - startDay + 1;
//Honor the maximum, if set
if (maxDays >= 0) {
days = Math.min(days, maxDays);
}
return days;
}
如果您需要日期之间的天数（不包括后一个日期），只需+ 1在您看到endDay - startDay + 1.
IT小君 另一种方式：
public static int numberOfDaysBetweenDates(Calendar fromDay, Calendar toDay) {
fromDay = calendarStartOfDay(fromDay);
toDay = calendarStartOfDay(toDay);
long from = fromDay.getTimeInMillis();
long to = toDay.getTimeInMillis();
return (int) TimeUnit.MILLISECONDS.toDays(to - from);
}
使用这些功能
public static int getDateDifference(int previousYear, int previousMonthOfYear, int previousDayOfMonth, int nextYear, int nextMonthOfYear, int nextDayOfMonth, int differenceToCount){
// int differenceToCount = can be any of the following
// Calendar.MILLISECOND;
// Calendar.SECOND;
// Calendar.MINUTE;
// Calendar.HOUR;
// Calendar.DAY_OF_MONTH;
// Calendar.MONTH;
// Calendar.YEAR;
// Calendar.----
Calendar previousDate = Calendar.getInstance();
previousDate.set(Calendar.DAY_OF_MONTH, previousDayOfMonth);
// month is zero indexed so month should be minus 1
previousDate.set(Calendar.MONTH, previousMonthOfYear);
previousDate.set(Calendar.YEAR, previousYear);
Calendar nextDate = Calendar.getInstance();
nextDate.set(Calendar.DAY_OF_MONTH, previousDayOfMonth);
// month is zero indexed so month should be minus 1
nextDate.set(Calendar.MONTH, previousMonthOfYear);
nextDate.set(Calendar.YEAR, previousYear);
return getDateDifference(previousDate,nextDate,differenceToCount);
}
public static int getDateDifference(Calendar previousDate,Calendar nextDate,int differenceToCount){
// int differenceToCount = can be any of the following
// Calendar.MILLISECOND;
// Calendar.SECOND;
// Calendar.MINUTE;
// Calendar.HOUR;
// Calendar.DAY_OF_MONTH;
// Calendar.MONTH;
// Calendar.YEAR;
// Calendar.----
//raise an exception if previous is greater than nextdate.
if(previousDate.compareTo(nextDate)>0){
throw new RuntimeException("Previous Date is later than Nextdate");
}
int difference=0;
while(previousDate.compareTo(nextDate)<=0){
difference++;
previousDate.add(differenceToCount,1);
}
return difference;
}
public void dateDifferenceExample() {
// Set the date for both of the calendar instance
GregorianCalendar calDate = new GregorianCalendar(2012, 10, 02,5,23,43);
GregorianCalendar cal2 = new GregorianCalendar(2015, 04, 02);
// Get the represented date in milliseconds
long millis1 = calDate.getTimeInMillis();
long millis2 = cal2.getTimeInMillis();
// Calculate difference in milliseconds
long diff = millis2 - millis1;
// Calculate difference in seconds
long diffSeconds = diff / 1000;
// Calculate difference in minutes
long diffMinutes = diff / (60 * 1000);
// Calculate difference in hours
long diffHours = diff / (60 * 60 * 1000);
// Calculate difference in days
long diffDays = diff / (24 * 60 * 60 * 1000);
Toast.makeText(getContext(), ""+diffSeconds, Toast.LENGTH_SHORT).show();
}
我找到了一种非常简单的方法来做到这一点，这就是我在我的应用程序中使用的方法。
假设您在 Time 对象中有日期（或其他，我们只需要毫秒）：
Time date1 = initializeDate1(); //get the date from somewhere
Time date2 = initializeDate2(); //get the date from somewhere
long millis1 = date1.toMillis(true);
long millis2 = date2.toMillis(true);
long difference = millis2 - millis1 ;
//now get the days from the difference and that's it
long days = TimeUnit.MILLISECONDS.toDays(difference);
//now you can do something like
if(days == 7)
{
//do whatever when there's a week of difference
}
if(days >= 30)
{
//do whatever when it's been a month or more
}
最好的方法是使用Joda-Time，这是您将添加到项目中的非常成功的开源库。
String date1 = "2015-11-11";
String date2 = "2013-11-11";
DateTimeFormatter formatter = new DateTimeFormat.forPattern("yyyy-MM-dd");
DateTime d1 = formatter.parseDateTime(date1);
DateTime d2 = formatter.parseDateTime(date2);
long diffInMillis = d2.getMillis() - d1.getMillis();
Duration duration = new Duration(d1, d2);
int days = duration.getStandardDays();
int hours = duration.getStandardHours();
int minutes = duration.getStandardMinutes();
如果您使用的是Android Studio，则很容易添加 joda-time。在你的 build.gradle (app) 中：
dependencies {
compile 'joda-time:joda-time:2.4'
compile 'joda-time:joda-time:2.4'
compile 'joda-time:joda-time:2.2'
}
Date userDob = new SimpleDateFormat("yyyy-MM-dd").parse(dob);
Date today = new Date();
long diff = today.getTime() - userDob.getTime();
int numOfDays = (int) (diff / (1000 * 60 * 60 * 24));
int hours = (int) (diff / (1000 * 60 * 60));
int minutes = (int) (diff / (1000 * 60));
int seconds = (int) (diff / (1000));
不是一个可靠的方法，最好使用JodaTime
这是一个近似值...
要从字符串中解析日期，您可以使用
虽然，因为你确定日期格式......你也可以
Integer.parseInt()在它的子字符串上做以获得它们的数值。